Q. 194.4( 10 Votes )

# Find the values of k, if the points A (k + 1, 2k), B (3k, 2k + 3) and C (5k - 1, 5k) are collinear.

We know that, if three points are collinear, then the area of triangle formed by these points is zero.

Since, the points A (k + 1, 2k), B (3k, 2k + 3) and C (5k - 1, 5k) are collinear.

Then, area of ΔABC = 0

1/2 [x1 (y2 – y3) + x2 (y3 - y1 ) + x3 (y1 - y2 )] = 0

Here,

x1 = k + 1,

x2 = 3k

x3 = 5k – 1

And

y1 = 2k

y2 = 2k + 3

y3 = 5k

[(k + 1)(2k + 3 - 5k) + 3k(5k – 2k) + (5k – 1)(2k + 3)]

[(k + 1)( - 3k + 3) + 3k(3k) + (5k - 1)(2k – 2k – 3)] = 0

[ - 3k2 + 3k - 3k + 3 + 9k2 - 15k + 3] = 0

(6k2 – 15k + 6) = 0 (multiply by 2)

6k2 – 15k + 2 = 0

2k2 – 5k + 2 = 0

[Divide by 3]

2k2 – 4k – k + 2 = 0

2k (k – 2) – 1(k – 2) = 0

(k – 2)(2k – 1) = 0

If k – 2 = 0, then k = 2

If 2k – 1 = 0, then k= 1/2

k = 2, 1/2

Hence, the required values of k are 2 and 1/2.

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